BTGmoderatorDC wrote:When 900 is divided by positive integer d, the remainder is r. For some integer N > 5000, when N is divided by positive integer D, the remainder is R. Is R > d?

(1) r = 1

(2) D = 23
Source: Magoosh

Very nice conceptual problem!
$$d,D\,\,\, \ge 1\,\,\,;\,\,\,\,N\,\,\, \ge \,\,5001\,\,\,\,\,\,\left( {\,{\rm{ints}}\,} \right)$$
$$\left\{ \matrix{
900 = {Q_1} \cdot d + r\,\,\,\,\,\,\left[ {\,0 \le r < d\,\,,\,\,\,\,r\,\,{\mathop{\rm int}} \,} \right]\,\,\,\,\,\,\,\left( * \right) \hfill \cr
N = {Q_2} \cdot D + R\,\,\,\,\,\,\,\left[ {\,0 \le R < D\,\,,\,\,\,R\,\,{\mathop{\rm int}} \,} \right]\,\,\,\,\,\left( {**} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,R\,\,\mathop > \limits^? \,\,d$$

The BIFURCATION of each statement alone will be omitted because it is easy (also intuitively).
(It is important, though. If you feel you need to do it, do not hesitate and hands-on!)

$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,r = 1\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{Q_1} \cdot d\, = 899 = 29 \cdot 31\,\,\,\left( {{\rm{primes}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,d = \,\,29\,,\,31\,\,{\rm{or}}\,\,899\,\,\,\left( {d > r = 1} \right)\,\,\,\,\, \hfill \cr
D = 23\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,0 \le R < 23\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,R\,\,\mathop > \limits^? \,\,d\,\,\, \ldots \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

P.S.: Jay´s approach to factorize 899 is excellent. Do not miss it (post below)!